Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 100020;
int n, m, q;
int f[Maxn][19], h[Maxn];
vector<int> has[Maxn][19];
vector<int> peo[Maxn];
vector<int> G[Maxn];
vector<int> operator+(const vector<int>& V1, const vector<int>& V2) {
  vector<int> ret = V1;
  for (int i = 0; i < V2.size(); i++) ret.push_back(V2[i]);
  sort(ret.begin(), ret.end());
  if (ret.size() > 10) ret.resize(10);
  return ret;
}
void dfs(int u, int p) {
  f[u][0] = p;
  sort(has[u][0].begin(), has[u][0].end());
  if (has[u][0].size() > 10) has[u][0].resize(10);
  for (int i = 0; f[u][i]; i++) {
    f[u][i + 1] = f[f[u][i]][i];
    has[u][i + 1] = has[u][i] + has[f[u][i]][i];
  }
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == p) continue;
    h[v] = h[u] + 1;
    dfs(v, u);
  }
}
vector<int> query(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  int del = h[u] - h[v];
  vector<int> rep;
  for (int i = 0; del; i++, del >>= 1) {
    if (del & 1) {
      rep = rep + has[u][i];
      u = f[u][i];
    }
  }
  for (int i = 17; i >= 0; i--) {
    if (f[u][i] != f[v][i]) {
      rep = rep + has[u][i];
      u = f[u][i];
      rep = rep + has[v][i];
      v = f[v][i];
    }
  }
  if (u == v)
    rep = rep + has[u][0];
  else {
    rep = rep + has[u][1];
    rep = rep + has[v][0];
  }
  return rep;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    has[x][0].push_back(i);
  }
  dfs(1, 0);
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    vector<int> rep = query(u, v);
    if (rep.size() > a) rep.resize(a);
    printf("%d", rep.size());
    for (int i = 0; i < rep.size(); i++) printf(" %d", rep[i]);
    puts("");
  }
}